The section formula and the midpoint

Sup­pose a road runs straight between two vil­lages and a bus stop is to be built one third of the way along. Where exactly is it? The sec­tion for­mula answers ques­tions like this: it gives the coor­di­nates of the point that divides a seg­ment in any ratio. Its sim­plest case, the mid­point for­mula, is used con­stantly. This les­son derives both and uses them to find divid­ing points, miss­ing end­points and unknown ratios.

The sec­tion for­mula#

A point $P$ on the seg­ment $AB$ divides it inter­nally in the ratio $m : n$ if $AP : PB = m : n$.

Sec­tion for­mula. If $A(x_1, y_1)$ and $B(x_2, y_2)$, the point $P$ divid­ing $AB$ inter­nally in the ratio $m : n$ is

$$\displaystyle P\left(\frac{mx_2 + nx_1}{m + n}, \; \frac{my_2 + ny_1}{m + n}\right)$$

Notice the cross pat­tern: $m$ (the part next to $A$) mul­ti­plies the coor­di­nates of $B$, and $n$ mul­ti­plies the coor­di­nates of $A$.

Deriva­tion. Let $P$ be $(x, y)$. Draw hor­i­zon­tal lines through $A$ and $P$, and ver­ti­cal lines through $P$ and $B$, to make two right tri­an­gles: one with hypotenuse $AP$ and legs $x - x_1$ and $y - y_1$, the other with hypotenuse $PB$ and legs $x_2 - x$ and $y_2 - y$. These tri­an­gles have equal angles (their sides are par­al­lel), so they are sim­i­lar, and their sides are in the ratio $AP : PB = m : n$:

$$\displaystyle \frac{x - x_1}{x_2 - x} = \frac{m}{n} \implies n(x - x_1) = m(x_2 - x) \implies x = \frac{mx_2 + nx_1}{m + n}$$

The same steps give $y$.

Exam­ple 1#

Find the point that divides the seg­ment join­ing $A(1, -2)$ and $B(7, 4)$ inter­nally in the ratio $1 : 2$.

Here $m = 1$, $n = 2$.

$$\displaystyle x = \frac{1 \times 7 + 2 \times 1}{1 + 2} = \frac{9}{3} = 3, \qquad y = \frac{1 \times 4 + 2 \times (-2)}{3} = \frac{0}{3} = 0$$

The point is $(3, 0)$. It is one third of the way from $A$ to $B$.

The mid­point for­mula#

The mid­point divides a seg­ment in the ratio $1 : 1$. Putting $m = n = 1$:

$$\displaystyle \text{Midpoint of } AB = \left(\frac{x_1 + x_2}{2}, \; \frac{y_1 + y_2}{2}\right)$$

The mid­point's coor­di­nates are the aver­ages of the end­points' coor­di­nates.

Exam­ple 2#

Find the mid­point of the seg­ment join­ing $(-3, 5)$ and $(7, -1)$.

$$\displaystyle \left(\frac{-3 + 7}{2}, \; \frac{5 + (-1)}{2}\right) = (2, 2)$$

Exam­ple 3#

The mid­point of $AB$ is $M(2, 3)$ and $A$ is $(-1, 7)$. Find $B$.

Let $B = (x, y)$. Then $\displaystyle \frac{-1 + x}{2} = 2$, so $x = 5$, and $\displaystyle \frac{7 + y}{2} = 3$, so $y = -1$. So $B = (5, -1)$.

A quick way to remem­ber it: $B = 2M - A$, coor­di­nate by coor­di­nate.

Find­ing the ratio#

When the divid­ing point is known and the ratio is not, call the ratio $k : 1$. Then $\displaystyle P = \left(\frac{kx_2 + x_1}{k + 1}, \frac{ky_2 + y_1}{k + 1}\right)$. Use whichever coor­di­nate is eas­ier to find $k$, and check with the other.

Exam­ple 4#

In what ratio does the point $P(0, 1)$ divide the seg­ment join­ing $A(-2, 5)$ and $B(4, -7)$?

Using the x-coor­di­nates:

$$\displaystyle \frac{4k + (-2)}{k + 1} = 0 \implies 4k - 2 = 0 \implies k = \frac{1}{2}$$

So the ratio is $\displaystyle \frac{1}{2} : 1 = 1 : 2$. Check with $y$: $\displaystyle \frac{\frac{1}{2}(-7) + 5}{\frac{3}{2}} = \frac{1.5}{1.5} = 1$. Cor­rect.

Exam­ple 5#

In what ratio does the x-axis divide the seg­ment join­ing $(2, -3)$ and $(5, 6)$? Find the point of divi­sion.

Every point on the x-axis has y-coor­di­nate $0$:

$$\displaystyle \frac{6k + (-3)}{k + 1} = 0 \implies k = \frac{1}{2}$$

The ratio is $1 : 2$, and the point is $\displaystyle \left(\frac{1 \times 5 + 2 \times 2}{3}, 0\right) = (3, 0)$.

Points of tri­sec­tion#

The two points that divide a seg­ment into three equal parts are its points of tri­sec­tion. They divide it in the ratios $1 : 2$ and $2 : 1$.

Exam­ple 6#

Find the points of tri­sec­tion of the seg­ment join­ing $A(2, -2)$ and $B(-7, 4)$.

Ratio $1 : 2$:

$$\displaystyle \left(\frac{1 \times (-7) + 2 \times 2}{3}, \; \frac{1 \times 4 + 2 \times (-2)}{3}\right) = (-1, 0)$$

Ratio $2 : 1$:

$$\displaystyle \left(\frac{2 \times (-7) + 1 \times 2}{3}, \; \frac{2 \times 4 + 1 \times (-2)}{3}\right) = (-4, 2)$$

Prac­tice#

1. Find the point divid­ing the seg­ment join­ing $(-1, 7)$ and $(4, -3)$ inter­nally in the ratio $2 : 3$.
2. Find the mid­point of the seg­ment join­ing $(4, -6)$ and $(-2, 8)$.
3. The mid­point of $AB$ is $(1, 2)$ and $A$ is $(3, -4)$. Find $B$.
4. Find the points of tri­sec­tion of the seg­ment join­ing $(1, -3)$ and $(7, 6)$.
5. In what ratio does the y-axis divide the seg­ment join­ing $(5, -6)$ and $(-1, -4)$? Find the point of divi­sion.
6. In what ratio does $(0, 3)$ divide the seg­ment join­ing $(-4, 1)$ and $(8, 7)$?
7. If $(p, 2)$ is the mid­point of the seg­ment join­ing $(3, 6)$ and $(7, -2)$, find $p$.
8. Find the points that divide the seg­ment from $(0, 0)$ to $(9, 12)$ into three equal parts.

Answers#